Mixing
Sequences and geometric progressions
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Any sequence of natural numbers that contains an infinite arithmetic progression (AP) must have a positive lower density and this alone rules out many candidates (squares, primes, etc.). On the other hand, a sequence can be quite dense without containing any geometric progressions (GP's) at all (e.g. squarefree numbers). Given any naturally occurring sequence one can ask for the longest GP it contains, if any. Some yield to a direct approach or to a
little theory (shifted two powers $2^n - 1$, Fibonacci numbers) while others seem less cooperative (shifted primes $p - 1$, factorials). Three specific instances follow although the general program (take any sequence & find its GP's) is obviously open-ended. $,$ To avoid trivialities, all progressions are assumed to have length $ge 3$.
Questions:
(1) It is known that the squares $n^2$ do not contain any $4$ term AP. Of course, they do have infinite GP's (like $4^n$) but the status of shifted squares $n^2 + 1$ [OEIS A002522] seems less clear. We at least have the three terms $2, 10, 50$ but is there any plausible upper bound to the length of a GP here? $;$ [Presumably, it is safe to conjecture that no polynomial value set $f(n), n=1,2,3,...$ contains an infinite GP as long as f is, say, irreducible of degree $ge 2$].
(2) The factorials $n$! clearly do not contain any infinite GP and even the existence of short GP's would be a surprise. This would entail having equal products of consecutive integers. Although this can happen ($5cdot6cdot7 = 14cdot15$), additionally we would need the blocks to be consecutive (a very tall order).$;$ Can GP's be ruled out for $n!$ altogether or, if not, at least those having length $ge 4$?
(3) Although it's an easy observation that the primes don't contain an infinite
AP, it is a difficult and recent result that they do contain arbitrarily long
AP's. Turning to the question of GP's, since it is clear that they don't exist in the primes themselves, we consider the shifted sequence $p - 1$ [A006093] instead. Unlike the previous two cases, GP's abound here. Two examples are $2, 6, 18$ and $16, 40, 100, 250$ (with common ratio = $5/2$). $;$ Should we expect an analogous result here, i.e. that the sequence $p - 1$ harbors arbitrarily long geometric progressions? $;$ [It can be shown that there are no infinite GP's].
Thanks.
number-theory geometric-progressions
edited Jul 18 at 19:56
an4s
2,0382417
asked Jul 18 at 19:37
user2052
1,089611
add a comment |Â
up vote
0
down vote
favorite
Any sequence of natural numbers that contains an infinite arithmetic progression (AP) must have a positive lower density and this alone rules out many candidates (squares, primes, etc.). On the other hand, a sequence can be quite dense without containing any geometric progressions (GP's) at all (e.g. squarefree numbers). Given any naturally occurring sequence one can ask for the longest GP it contains, if any. Some yield to a direct approach or to a
little theory (shifted two powers $2^n - 1$, Fibonacci numbers) while others seem less cooperative (shifted primes $p - 1$, factorials). Three specific instances follow although the general program (take any sequence & find its GP's) is obviously open-ended. $,$ To avoid trivialities, all progressions are assumed to have length $ge 3$.
Questions:
(1) It is known that the squares $n^2$ do not contain any $4$ term AP. Of course, they do have infinite GP's (like $4^n$) but the status of shifted squares $n^2 + 1$ [OEIS A002522] seems less clear. We at least have the three terms $2, 10, 50$ but is there any plausible upper bound to the length of a GP here? $;$ [Presumably, it is safe to conjecture that no polynomial value set $f(n), n=1,2,3,...$ contains an infinite GP as long as f is, say, irreducible of degree $ge 2$].
(2) The factorials $n$! clearly do not contain any infinite GP and even the existence of short GP's would be a surprise. This would entail having equal products of consecutive integers. Although this can happen ($5cdot6cdot7 = 14cdot15$), additionally we would need the blocks to be consecutive (a very tall order).$;$ Can GP's be ruled out for $n!$ altogether or, if not, at least those having length $ge 4$?
(3) Although it's an easy observation that the primes don't contain an infinite
AP, it is a difficult and recent result that they do contain arbitrarily long
AP's. Turning to the question of GP's, since it is clear that they don't exist in the primes themselves, we consider the shifted sequence $p - 1$ [A006093] instead. Unlike the previous two cases, GP's abound here. Two examples are $2, 6, 18$ and $16, 40, 100, 250$ (with common ratio = $5/2$). $;$ Should we expect an analogous result here, i.e. that the sequence $p - 1$ harbors arbitrarily long geometric progressions? $;$ [It can be shown that there are no infinite GP's].
Thanks.
number-theory geometric-progressions
edited Jul 18 at 19:56
an4s
2,0382417
asked Jul 18 at 19:37
user2052
1,089611
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Any sequence of natural numbers that contains an infinite arithmetic progression (AP) must have a positive lower density and this alone rules out many candidates (squares, primes, etc.). On the other hand, a sequence can be quite dense without containing any geometric progressions (GP's) at all (e.g. squarefree numbers). Given any naturally occurring sequence one can ask for the longest GP it contains, if any. Some yield to a direct approach or to a
little theory (shifted two powers $2^n - 1$, Fibonacci numbers) while others seem less cooperative (shifted primes $p - 1$, factorials). Three specific instances follow although the general program (take any sequence & find its GP's) is obviously open-ended. $,$ To avoid trivialities, all progressions are assumed to have length $ge 3$.
Questions:
(1) It is known that the squares $n^2$ do not contain any $4$ term AP. Of course, they do have infinite GP's (like $4^n$) but the status of shifted squares $n^2 + 1$ [OEIS A002522] seems less clear. We at least have the three terms $2, 10, 50$ but is there any plausible upper bound to the length of a GP here? $;$ [Presumably, it is safe to conjecture that no polynomial value set $f(n), n=1,2,3,...$ contains an infinite GP as long as f is, say, irreducible of degree $ge 2$].
(2) The factorials $n$! clearly do not contain any infinite GP and even the existence of short GP's would be a surprise. This would entail having equal products of consecutive integers. Although this can happen ($5cdot6cdot7 = 14cdot15$), additionally we would need the blocks to be consecutive (a very tall order).$;$ Can GP's be ruled out for $n!$ altogether or, if not, at least those having length $ge 4$?
(3) Although it's an easy observation that the primes don't contain an infinite
AP, it is a difficult and recent result that they do contain arbitrarily long
AP's. Turning to the question of GP's, since it is clear that they don't exist in the primes themselves, we consider the shifted sequence $p - 1$ [A006093] instead. Unlike the previous two cases, GP's abound here. Two examples are $2, 6, 18$ and $16, 40, 100, 250$ (with common ratio = $5/2$). $;$ Should we expect an analogous result here, i.e. that the sequence $p - 1$ harbors arbitrarily long geometric progressions? $;$ [It can be shown that there are no infinite GP's].
Thanks.
number-theory geometric-progressions
edited Jul 18 at 19:56
an4s
2,0382417
asked Jul 18 at 19:37
user2052
1,089611
Any sequence of natural numbers that contains an infinite arithmetic progression (AP) must have a positive lower density and this alone rules out many candidates (squares, primes, etc.). On the other hand, a sequence can be quite dense without containing any geometric progressions (GP's) at all (e.g. squarefree numbers). Given any naturally occurring sequence one can ask for the longest GP it contains, if any. Some yield to a direct approach or to a
little theory (shifted two powers $2^n - 1$, Fibonacci numbers) while others seem less cooperative (shifted primes $p - 1$, factorials). Three specific instances follow although the general program (take any sequence & find its GP's) is obviously open-ended. $,$ To avoid trivialities, all progressions are assumed to have length $ge 3$.
Questions:
(1) It is known that the squares $n^2$ do not contain any $4$ term AP. Of course, they do have infinite GP's (like $4^n$) but the status of shifted squares $n^2 + 1$ [OEIS A002522] seems less clear. We at least have the three terms $2, 10, 50$ but is there any plausible upper bound to the length of a GP here? $;$ [Presumably, it is safe to conjecture that no polynomial value set $f(n), n=1,2,3,...$ contains an infinite GP as long as f is, say, irreducible of degree $ge 2$].
(2) The factorials $n$! clearly do not contain any infinite GP and even the existence of short GP's would be a surprise. This would entail having equal products of consecutive integers. Although this can happen ($5cdot6cdot7 = 14cdot15$), additionally we would need the blocks to be consecutive (a very tall order).$;$ Can GP's be ruled out for $n!$ altogether or, if not, at least those having length $ge 4$?
(3) Although it's an easy observation that the primes don't contain an infinite
AP, it is a difficult and recent result that they do contain arbitrarily long
AP's. Turning to the question of GP's, since it is clear that they don't exist in the primes themselves, we consider the shifted sequence $p - 1$ [A006093] instead. Unlike the previous two cases, GP's abound here. Two examples are $2, 6, 18$ and $16, 40, 100, 250$ (with common ratio = $5/2$). $;$ Should we expect an analogous result here, i.e. that the sequence $p - 1$ harbors arbitrarily long geometric progressions? $;$ [It can be shown that there are no infinite GP's].
Thanks.
number-theory geometric-progressions
edited Jul 18 at 19:56
an4s
2,0382417
asked Jul 18 at 19:37
user2052
1,089611
edited Jul 18 at 19:56
an4s
2,0382417
edited Jul 18 at 19:56
an4s
2,0382417
edited Jul 18 at 19:56
an4s
2,0382417
2,0382417
asked Jul 18 at 19:37
user2052
1,089611
asked Jul 18 at 19:37
user2052
1,089611
asked Jul 18 at 19:37
user2052
1,089611
1,089611
add a comment |Â
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855925%2fsequences-and-geometric-progressions%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password